Microsoft Word - Cengel and Boles TOC _2-03-05_.doc

14–3
DEW-POINT TEMPERATURE

If you live in a humid area, you are probably used to waking up most summer
mornings and finding the grass wet. You know it did not rain the night before.
So what happened? Well, the excess moisture in the air simply condensed on
the cool surfaces, forming what we call dew.In summer, a considerable
amount of water vaporizes during the day. As the temperature falls during the

Chapter 14 | 721

Analysis (a) The partial pressure of dry air can be determined from Eq. 14–2:

where

Thus,

(b) The specific humidity of air is determined from Eq. 14–8:

(c) The enthalpy of air per unit mass of dry air is determined from

Eq. 14–12:

The enthalpy of water vapor (2546.5 kJ/kg) could also be determined from

the approximation given by Eq. 14–4:

which is almost identical to the value obtained from Table A–4.

(d) Both the dry air and the water vapor fill the entire room completely.

Therefore, the volume of each gas is equal to the volume of the room:

The masses of the dry air and the water vapor are determined from the ideal-

gas relation applied to each gas separately:

The mass of the water vapor in the air could also be determined from

Eq. 14–6:

mvvma 1 0.0152 21 85.61 kg 2 1.30 kg

mv

PvVv

RvT



1 2.38 kPa 21 75 m^32

1 0.4615 kPa#m^3 >kg#K 21 298 K 2

1.30 kg

ma

PaVa

RaT



1 97.62 kPa 21 75 m^32

1 0.287 kPa#m^3 >kg#K 21 298 K 2

85.61 kg

VaVvVroom 1 5 m 21 5 m 21 3 m 2 75 m^3

hg @ 25°C
2500.91.82 1252 2546.4 kJ>kg

63.8 kJ/kg dry air

 1 1.005 kJ>kg#°C 21 25°C 2  1 0.0152 21 2546.5 kJ>kg 2

hhavhv
cpTvhg

v

0.622Pv

P
Pv



1 0.622 21 2.38 kPa 2

1100 
2.38 2 kPa

0.0152 kg H 2 O/kg dry air

Pa 1100 
2.38 2 kPa97.62 kPa

PvfPgfPsat @ 25°C 1 0.75 21 3.1698 kPa 2 2.38 kPa

PaP
Pv